GCSE · Geometry and measures

Sine and Cosine Rules Higher

SOH CAH TOA only works when there's a right angle. Most triangles in the real world (a field, a roof truss, a ship's route) don't have one. Two rules, both built by cutting a triangle into right-angled pieces, let you find any side or angle of any triangle, and a third formula gives its area.

GCSE subject content G22 (Higher): know and apply the sine rule and cosine rule to find unknown lengths and angles. G23 (Higher): know and apply Area = ½ab sin C to calculate the area, sides or angles of any triangle.

1 · Labelling any triangle Higher

You need right-angled trigonometry first (see Trigonometry and exact values, or the KS3 page Trigonometric ratios). For a general triangle we use a standard labelling:

A side and its opposite angle make a pair (same colour below). Spotting pairs is the key to choosing a rule. Drag the corners: the biggest angle is always opposite the longest side.

Pair a, A
Pair b, B
Pair c, C

Lengths are in cm. Angles always add up to .

2 · The sine rule: where it comes from Higher

Cut the triangle into two right-angled triangles by dropping a perpendicular (a line at 90°) from C to the line AB. Call its length h. Then write h in two ways, once from each right-angled triangle. Step through, and drag the corners to check that it works for every shape, even when the perpendicular lands outside the triangle.

Sine rule (for finding a side): a/sin A = b/sin B = c/sin C
Upside down (for finding an angle): sin A/a = sin B/b = sin C/c

You only ever use two of the three fractions at once. You need one complete pair (a side and its opposite angle) plus one more piece of information.

3 · Using the sine rule Higher

To find a side, put the unknown side on top: b/sin B = a/sin A, then multiply both sides by sin B. To find an angle, put the sines on top: sin B/b = sin A/a, multiply by b, then use sin⁻¹. If you're given two angles, the third one is 180° minus the other two; you might need it to make a pair.

4 · The ambiguous case Higher

Here's a surprise. Suppose you know angle A, side b and side a (which is opposite A). Picture side a as a stick hinged at C, swinging until it hits the base line. Sometimes it can land in two places, giving two different triangles that both fit the information. That's because sin x = sin(180° − x): for example sin 50° = sin 130°. So when the sine rule gives you an angle, there may be a second, obtuse answer: 180° minus the first.

When to worry. The ambiguous case only happens when you use the sine rule to find an angle, and the side opposite the known angle is the shorter of the two given sides. If the question says the angle is obtuse, or the diagram shows it, take 180° minus your calculator answer. Otherwise, check whether 180° minus it still leaves room in the triangle (the angles must add to 180°).

5 · The cosine rule: Pythagoras with a correction Higher

If you know two sides and the angle between them, there's no complete pair, so the sine rule can't start. Pythagoras says c² = a² + b² when C = 90°. For other angles, c² needs a correction. Make C smaller than 90° and the triangle closes up, so c is shorter than Pythagoras says; make C bigger and c is longer. The correction is exactly 2ab cos C.

Cosine rule (for a side): c² = a² + b² − 2ab cos C
Rearranged (for an angle): cos C = (a² + b² − c²) ÷ 2ab
Why is the correction 2ab cos C? (algebraic proof)
Drop a perpendicular h from B onto side CA, landing at D. Let CD = x, so DA = b − x.
In the right-angled triangle BCD: x = a cos C and h² = a² − x².
In the right-angled triangle BDA: c² = h² + (b − x)² = a² − x² + b² − 2bx + x² = a² + b² − 2bx.
Replace x by a cos C: c² = a² + b² − 2ab cos C. When C = 90°, cos C = 0 and this is Pythagoras.

Practise both ways round. The commonest error is BIDMAS: in a² + b² − 2ab cos C, work out 2ab cos C first and subtract it. Don't do a² + b² − 2ab and then multiply by cos C.

6 · Area = ½ab sin C Higher

You know area = ½ × base × height. Take a as the base. The height is the perpendicular from A to the line BC, and in the right-angled triangle it makes with side b, height = b sin C. So

area = ½ × a × b sin C

Use it with two sides and the angle between them (the included angle). The letters can change, e.g. ½bc sin A; the pattern is "two sides and the angle where they meet". Drag the corners: both calculations always agree.

7 · Which rule? Higher

The hardest part of an exam question is often choosing. Work through the flowchart with a question in mind.

Now practise choosing. Each triangle shows what you know; the question mark is what you want.

8 · Bearings and real problems Higher

A ship sails from A to B, then turns and sails to C. How far is C from A, and on what bearing? (Bearings are measured clockwise from north as three figures; see the Bearings page.) The two legs and the angle between them at B give two sides and the included angle: a cosine rule question. The hard part is finding angle ABC, using the parallel north lines.

Worked examples

Sine rule, finding a side: A = 40°, B = 65°, a = 8 cm. Find b. Higher
Pairs: a = 8 with A = 40° is a complete pair. b with B = 65° is the pair we want. → sine rule.
b / sin 65° = 8 / sin 40°
b = 8 × sin 65° ÷ sin 40° = 8 × 0.906307… ÷ 0.642787… = 11.279… cm
Answer: b = 11.3 cm (3 s.f.). Sense check: B is bigger than A, so b should be longer than a ✓. (Upside down gives 5.67 cm, shorter than a: wrong.)
Sine rule, finding an angle: a = 9 cm, b = 7 cm, A = 72°. Find B. Higher
Pairs: a, A complete; want B, and we know b. → sine rule, angles on top.
sin B / 7 = sin 72° / 9 → sin B = 7 × sin 72° ÷ 9 = 0.739710…
B = sin⁻¹(0.739710…) = 47.706…° → B = 47.7° (1 d.p.)
Could B be 180° − 47.7° = 132.3°? Then A + B = 72° + 132.3° = 204.3°, more than 180°. Impossible, so there's only one answer. (Also b is shorter than a, so B must be smaller than A.)
The ambiguous case: A = 35°, a = 6 cm, b = 9 cm. Find the two possible values of B. Higher
sin B = 9 × sin 35° ÷ 6 = 0.860364…
B = sin⁻¹(0.860364…) = 59.4° (1 d.p.)
Second solution: 180° − 59.357…° = 120.6°. Check it fits: 35° + 120.6° = 155.6°, less than 180°, leaving C = 24.4°. Both triangles exist.
This happens because a (6) is shorter than b (9) but longer than the height b sin A = 5.16: the swinging side can land in two places.
Cosine rule, finding a side: b = 7 cm, c = 10 cm, A = 50°. Find a. Higher
Two sides and the included angle, no complete pair → cosine rule.
a² = b² + c² − 2bc cos A = 7² + 10² − 2 × 7 × 10 × cos 50°
a² = 49 + 100 − 140 × 0.642787… = 149 − 89.990… = 59.0097…
a = √59.0097… = 7.681… → a = 7.68 cm (3 s.f.)
Trap: (49 + 100 − 140) × cos 50° = 5.79 is the BIDMAS error. And don't forget the square root: 59.0 is a², not a.
Cosine rule, finding an angle: a triangle has sides 5 cm, 7 cm and 9 cm. Find its largest angle. Higher
Largest angle is opposite the longest side (9 cm). Call it C, so c = 9, a = 5, b = 7.
cos C = (a² + b² − c²) ÷ 2ab = (25 + 49 − 81) ÷ (2 × 5 × 7) = −7 ÷ 70 = −0.1
C = cos⁻¹(−0.1) = 95.739…° → 95.7° (1 d.p.)
A negative cosine means an obtuse angle. Unlike the sine rule, cos⁻¹ never gives an ambiguous answer.
Area: a = 8 cm, b = 11 cm, C = 40°. Find the area of the triangle. Higher
C is between sides a and b (the included angle) ✓.
Area = ½ab sin C = ½ × 8 × 11 × sin 40° = 44 × 0.642787… = 28.282…
Answer: 28.3 cm² (3 s.f.)
Bearings: a ship sails 15 km on a bearing of 040° from A to B, then 9 km on a bearing of 120° to C. Find the distance AC and the bearing of C from A. Higher
Angle ABC. At B, the bearing back to A is 040° + 180° = 220°. The ship then heads on 120°. The angle between those two directions is 220° − 120° = 100°.
Cosine rule: AC² = 15² + 9² − 2 × 15 × 9 × cos 100° = 225 + 81 + 46.884… = 352.885…, so AC = 18.785… = 18.8 km (3 s.f.).
Angle BAC (sine rule): sin BAC = 9 × sin 100° ÷ 18.785… = 0.471…, so BAC = 28.15…° (acute, as it's opposite the shortest side).
Bearing: C is clockwise of the first leg, so 040° + 28.15…° = 068.15…° → 068° to the nearest degree.

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